# FP2: Taylors series

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#3

(Original post by

do I use this formula and make a = 0?

**Maths&physics**)do I use this formula and make a = 0?

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#4

(Original post by

any ideas how I'd tackle this?

**Maths&physics**)any ideas how I'd tackle this?

implies and , this means that

To get the fourth term, you can differentiate the ODE and see what is. The second term should be fairly obvious.

Last edited by RDKGames; 1 year ago

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(Original post by

You seek a truncated expansion of , where it can be expanded in powers of as:

implies and , this means that

To get the fourth term, you can differentiate the ODE and see what is. The second term should be fairly obvious.

**RDKGames**)You seek a truncated expansion of , where it can be expanded in powers of as:

implies and , this means that

To get the fourth term, you can differentiate the ODE and see what is. The second term should be fairly obvious.

(Original post by

Presume so. The initial conditions give you the first two terms. The ode (rearranged) would give you the third term (t^2) and you could differentiate the ode again, to get the last (t^3) term.

**mqb2766**)Presume so. The initial conditions give you the first two terms. The ode (rearranged) would give you the third term (t^2) and you could differentiate the ode again, to get the last (t^3) term.

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#6

Sorry, the differential equation. Just differentiate it with respect to x again.

ODE is Ordinary Differential Equation - you don't need to know the term, but can look it up if you want.

ODE is Ordinary Differential Equation - you don't need to know the term, but can look it up if you want.

(Original post by

what is ODE?

**Maths&physics**)what is ODE?

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#7

(Original post by

what is ODE?

**Maths&physics**)what is ODE?

That is really not a question you want to be asking at this stage...

ODE = Ordinary Differential Equation

I.e. the equation with the derivative that they give you!

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(Original post by

Oh dear...

That is really not a question you want to be asking at this stage...

ODE = Ordinary Differential Equation

I.e. the equation with the derivative that they give you!

**RDKGames**)Oh dear...

That is really not a question you want to be asking at this stage...

ODE = Ordinary Differential Equation

I.e. the equation with the derivative that they give you!

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#9

(Original post by

is that a term in the specification because I've never heard it before.

**Maths&physics**)is that a term in the specification because I've never heard it before.

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#10

**Maths&physics**)

is that a term in the specification because I've never heard it before.

But TBH, I would be very surprised (and disappointed) if the resources you use for this theory do not abbreviate the topic to ODE because it's an extremely standard acronym.

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(Original post by

No.

**mqb2766**)No.

**RDKGames**)

Oh dear...

That is really not a question you want to be asking at this stage...

ODE = Ordinary Differential Equation

I.e. the equation with the derivative that they give you!

I've got values for x but what would dx/dt =?

Last edited by Maths&physics; 1 year ago

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#12

(Original post by

I got for the third term... (d^3)y/d(t^2) = + (dx/dt)[1-sin(x)]dx/dt = 0

I've got values for x but what would dx/dt =?

**Maths&physics**)I got for the third term... (d^3)y/d(t^2) = + (dx/dt)[1-sin(x)]dx/dt = 0

I've got values for x but what would dx/dt =?

Edit that expression has gone a bit wonky?

Last edited by mqb2766; 1 year ago

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#13

**Maths&physics**)

I got for the third term... (d^3)y/d(t^2) = + (dx/dt)[1-sin(x)]dx/dt = 0

I've got values for x but what would dx/dt =?

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(Original post by

Its evaulated at t=0. You're given the value in the initial condition.

Edit that expression has gone a bit wonky?

**mqb2766**)Its evaulated at t=0. You're given the value in the initial condition.

Edit that expression has gone a bit wonky?

(Original post by

That equation looks dodgy... how did you arrive at that??

**RDKGames**)That equation looks dodgy... how did you arrive at that??

????

ok, what do I do next? I don't understand how to get dx/dt?

I sub x = 0, and I get: (d^3)y/d(t^3) = -(dx/dt)

Last edited by Maths&physics; 1 year ago

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#15

(Original post by

I meant: (d^3)y/d(t^3) = + (dx/dt)[1-sin(x)]dx/dt = 0

????

ok, what do I do next?

**Maths&physics**)I meant: (d^3)y/d(t^3) = + (dx/dt)[1-sin(x)]dx/dt = 0

????

ok, what do I do next?

(a) what are you actually doing to this equation...?

(b) why do you have =0 there...?

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(Original post by

That's not right.

(a) what are you actually doing to this equation...?

(b) why do you have =0 there...?

**RDKGames**)That's not right.

(a) what are you actually doing to this equation...?

(b) why do you have =0 there...?

(d^3)y/d(t^3) = -(dx/dt)

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#17

(Original post by

(d^3)y/d(t^3) + (dx/dt)[1-sin(x)]dx/dt = 0

(d^3)y/d(t^3) = -(dx/dt)

**Maths&physics**)(d^3)y/d(t^3) + (dx/dt)[1-sin(x)]dx/dt = 0

(d^3)y/d(t^3) = -(dx/dt)

You should really lay it out properly.

When , we get:

You know every single value here apart from ... so rearrange for it.

Last edited by RDKGames; 1 year ago

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(Original post by

Nope...

When , we get:

You know every single value here apart from ... so rearrange for it.

**RDKGames**)Nope...

When , we get:

You know every single value here apart from ... so rearrange for it.

that's what I meant!

(d^3)x/d(t^3) + (dx/dt)[1-sin(x)]dx/dt = 0

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**mqb2766**)

Its evaulated at t=0. You're given the value in the initial condition.

Edit that expression has gone a bit wonky?

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